Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 9x}{x - 10} = \dfrac{23x - 49}{x - 10}$
Answer: Multiply both sides by $x - 10$ $ \dfrac{x^2 + 9x}{x - 10} (x - 10) = \dfrac{23x - 49}{x - 10} (x - 10)$ $ x^2 + 9x = 23x - 49$ Subtract $23x - 49$ from both sides: $ x^2 + 9x - (23x - 49) = 23x - 49 - (23x - 49)$ $ x^2 + 9x - 23x + 49 = 0$ $ x^2 - 14x + 49 = 0$ Factor the expression: $ (x - 7)(x - 7) = 0$ Therefore $x = 7$ The original expression is defined at $x = 7$ and $x = 7$, so there are no extraneous solutions.